Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{x + 8}{x + 3} \div \dfrac{x^2 + 14x + 48}{x^2 - 3x - 18} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{x + 8}{x + 3} \times \dfrac{x^2 - 3x - 18}{x^2 + 14x + 48} $ First factor out any common factors. $k = \dfrac{x + 8}{x + 3} \times \dfrac{x^2 - 3x - 18}{x^2 + 14x + 48} $ Then factor the quadratic expressions. $k = \dfrac {x + 8} {x + 3} \times \dfrac {(x + 3)(x - 6)} {(x + 8)(x + 6)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {(x + 8) \times (x + 3)(x - 6) } {(x + 3) \times (x + 8)(x + 6) } $ $k = \dfrac {(x + 3)(x - 6)(x + 8)} {(x + 8)(x + 6)(x + 3)} $ Notice that $(x + 8)$ and $(x + 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {(x + 3)(x - 6)\cancel{(x + 8)}} {\cancel{(x + 8)}(x + 6)(x + 3)} $ We are dividing by $x + 8$ , so $x + 8 \neq 0$ Therefore, $x \neq -8$ $k = \dfrac {\cancel{(x + 3)}(x - 6)\cancel{(x + 8)}} {\cancel{(x + 8)}(x + 6)\cancel{(x + 3)}} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $k = \dfrac {x - 6} {x + 6} $ $ k = \dfrac{x - 6}{x + 6}; x \neq -8; x \neq -3 $